Sketch inverse function (1 Viewer)

[YBK]

Hey could anyone please help me out in sketching:

y = 2sin^-1(x/pie)

Thanks!

 

[SoulSearcher]

Is that pi or pi * e?

Anyway, if it is y = 2sin^-1(x/pi), then the domain and range of that function should be
-pi < x < pi
-pi < y < pi

So it's your basic sin^-1x graph with the boundaries extended.

 

[Riviet]

I'm pretty sure he just meant pi. XD

 

[SoulSearcher]

Yeah, I was thinking he meant that as well

EDIT: Domain of sin^-1x is -1 < x < 1
Range of sin^-1x is -pi/2 < y < pi/2

I'm working on my memory here, can't be too sure if that's right.

 

[YBK]

ohh.. but how do we know it's in that domain?

and yeah i meant pi.. lol
must be thinking of pie...

 

[SoulSearcher]

-1 < x/pi < 1
= -pi < x < pi

Pie's good

 

[Riviet]

A systematic approach to the range:
Given y=2sin^-1(x/pi),
-pi/2 < sin^-1(x/pi) < pi/2
-pi < 2sin^-1(x/pi) < pi
.'. Range is -pi < y < pi

 

[YBK]

A systematic approach to the range:
Given y=2sin^-1(x/pi),
-pi/2 < sin^-1(x/pi) < pi/2
-pi < 2sin^-1(x/pi) < pi
.'. Range is -pi < y < pi

how'd you get the first step... ?

 

[Riviet]

how'd you get the first step... ?

That's the general range for the graph y=sin^-1x. The x/18 will just stretch/squash the graph horizontally and will do nothing to the range.

In general: sin^-1[f(x)] will basically always have the same basic range between pi/2 and -pi/2 inclusive.